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TopNewestBy Rational Root Theorem, there's no rational roots So if it can be factored, then it must be the product of two irreducible quadratic polynomials.

\[ (x^2+ax+b)(x^2+cx + d) \]

Because the sum of roots of this expression equals to 0, this motivates us to have \(a = - c\).

And because the product of roots is 100, then \(bd = 100 \).

So we have

\[ (x^2+ ax + b)(x^2 - ax + 100/b) \]

Expand and compare the coefficients gives \((a,b) = (-5,5), (5,20 ) \). – Pi Han Goh · 1 year, 6 months ago

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